Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+y &= 2 \\ 2x-y &= -3\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $-y = -2x-3$ Divide both sides by $-1$ to isolate $y$ $y = {2x + 3}$ Substitute this expression for $y$ in the first equation. $-8x+({2x + 3}) = 2$ $-8x + 2x + 3 = 2$ Simplify by combining terms, then solve for $x$ $-6x + 3 = 2$ $-6x = -1$ $x = \dfrac{1}{6}$ Substitute $\dfrac{1}{6}$ for $x$ back into the top equation. $-8( \dfrac{1}{6})+y = 2$ $-\dfrac{4}{3}+y = 2$ $y = \dfrac{10}{3}$ $y = \dfrac{10}{3}$ The solution is $\enspace x = \dfrac{1}{6}, \enspace y = \dfrac{10}{3}$.